AS June 2024 Paper 1 Q1
1 The quadratic equation \(x^2 + ax + b = 0\), where \(a\) and \(b\) are real constants, has a root \(2 - 3\mathrm{i}\).
(a) Write down the other root. [1]
(b) Hence or otherwise determine the values of \(a\) and \(b\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(2 + 3\mathrm{i}\) | B1 | 1.2 |
| [1] |
| Scheme | Marks | AO |
|---|---|---|
| \((x - 2 - 3\mathrm{i})(x - 2 + 3\mathrm{i}) = 0\) | M1 | 1.1 |
| \(\Rightarrow (x - 2)^2 + 9 = 0\) | A1 | 1.1 |
| \(\Rightarrow x^2 - 4x + 13 = 0\) [so \(a = -4\) and \(b = 13\)] | A1 | 1.1 |
| [3] |
Notes
M1: factors \(x - (2 + 3\mathrm{i})\) and \(x - (2 - 3\mathrm{i})\)
A1: or \(x^2 - 2x + 3\mathrm{i}x - 2x + 4 - 6\mathrm{i} - 3\mathrm{i}x + 6\mathrm{i} - 9\mathrm{i}^2\) or better
Alternative solution 1
| Scheme | Marks |
|---|---|
| \((2 - 3\mathrm{i})^2 + a(2 - 3\mathrm{i}) + b = 0\) | M1 |
| \(\Rightarrow -5 + 2a + b = 0,\ -12 - 3a = 0\) | A1 |
| \(\Rightarrow a = -4,\ b = 13\) | A1 |
M1: use of factor theorem
Alternative solution 2
| Scheme | Marks |
|---|---|
| \(\dfrac{-a \pm \sqrt{a^2 - 4b}}{2} = 2 \pm 3\mathrm{i}\) | M1 |
| \(a^2 - 4b = -36,\ -a/2 = 2\) | A1 |
| \(\Rightarrow a = -4,\ b = 13\) | A1 |
M1: use of quadratic formula
Alternative solution 3
| Scheme | Marks |
|---|---|
| \(a = -(\alpha + \beta)\) | M1 |
| \(= -4\) | A1 |
| \(b = \alpha\beta = (2 + 3\mathrm{i})(2 - 3\mathrm{i}) = 13\) | A1 |
M1: sum or product of roots formulae used, condone sign errors